Countable / Uncountable / Perfect Sets Question(s) tl;dr / Warning: wordy question, and I may have answered my own question right at the bottom. 
I just went down a rabbit hole of reading some Rudin's PMA Chapter 2 and thought about (in $\Re$) - condensation points, perfect sets, and I also came across the Cantor Set on wikipedia.
My question starts here:
Start by letting the set $ A $ have one point $ x \in \Re\ $.
Then we do the following process/task:
Keep adding points (that were not in $A$ previously), one by one, to $A$, so that every new point ends up being a limit point of $A$.
You could at this point ask, "How exactly are you adding new points?". To which I guess my answer is: "I guess there are many (uncountable) ways of doing this, but for the sake of the questions at the bottom, it doesn't matter". On second thought, maybe it does matter whether we restrict these points to a bounded subset of $\Re \ $ or not, but anyway...
Now, before reading up on perfect sets, I thought this set would be enumerable and therefore countable, because it appears to be a countable union of countable (limit) points.
However, now I think $A$ fits the definition of "perfect set", because
a) every point of $A$ is a limit point of $A$, and 
b)by the construction of $A$ (i.e. we only add points that contribute to making previous points into limit points), every limit point of $A$ is a point of $A$ (i.e. $A$ is closed).
Which is the definition of a perfect set (correct me if I'm wrong).
So, assuming the set $A$ was well-defined the way I defined it, the set $A$ must be enumerable (countable) because I'm adding points one by one. But also it is perfect - and therefore uncountable set? A contradiction! Which would mean that set construction "does not work". Why does my attempted set construction not work? It must be because of the "adding points one by one"? 
Also, does my attempted set construction have anything to do with nowhere dense perfect sets like the Cantor set?
My question(s) ends here.
My thoughts on an answer: With regards to the task being trying to make every point of $A$ a limit point of $A$, every time I add a new point, it creates an even harder task to overcome than before I added that point!
 A: If you start with one point, this set has no limit points, so we cannot add points one by one that are limit points of the previous ones. If we have any finite set this is closed, so has no limit points, and there is nothing to add. So concretely: what point do you add to $\{0\}$ such that $0$ is a limit point of the new set? You cannot.
If you start with a set like $\Bbb Q$, all of its points are limit points (it has no isolated points) but there are still uncountably many points left that are also a limit point of it but not yet in it, and you cannot reach them all in countably many steps. In $\Bbb R$ this is unavoidable: a set $A$ that is equal to its set of limit points $A'=A$, or a perfect set, must be uncountable (typical examples are $[0,1]$, $\Bbb R$ itself, or the Cantor set). 
A countable process as you imagine cannot start with any finite set, because at that stage there are no limit points to add. If you want to add points that are limit points, you can add a whole sequence of distinct points at the time, with the new limit point as its (sequential) limit, but then the points of that sequence you added are not yet limit points of the set at that stage and you have to add infinitely many new points to fix that etc. It's not going to be a simple matter of adding a point at the time. That way your set will remain countable and never be a perfect set. 
A: I think bof's comment is on the money. Consider the following example.
Let $\ A_1 = \{0,1\}.$
For each $\ i\in \mathbb{N},\ $ carry out the following procedure:
For each $\ x\in A_i,\ $ let $\ y_x = \min \{ d(x,z): z\in A_i\}.\ $ Then, the following definition makes sense. Define:
$$A_{i+1} = \bigcup_{x\in A_i} \left\{ x + \frac{1}{4}y_x: y_x = \min \{d(x,z): z\in A_i\} \right\}$$
$A_n\ $ is finite and well-defined for all $\ n\in\mathbb{N}.\ $ Therefore, we may construct a set
$$ X = \bigcup_{i\in\mathbb{N}} A_i.$$
If $\ x\in X,\ $ the $\ x\in A_i\ $ for some $\ i\in\mathbb{N},\ $ and is the limit point of  $\ \displaystyle\bigcup_{n\geq i+1} A_n\subset X,\ $ by the way $\ X\ $ was constructed. In other words, every member of $\ X\ $ is a limit point of $\ X.$
Furthermore, $\ X\ $ is the countable union of finite sets, and therefore $\ X\ $ is countable. Therefore $\ X\ $ is not perfect, i.e. there are limit points of $\ X\ $ that are not members of $\ X.\ $
My first reaction was this:

But what are these points? I don't think they exist...

But then I realised that such points do exist: $\ \displaystyle\sum_{k=1}^{\infty} \frac{1}{4^k} = \frac{1}{3} \not\in X.$
I guess the point is that there are countably many of these limit points of $\ X\ $ that are not in $\ X,\ $ otherwise we could just add these limit point to $\ X\ $ and $\ X\ $ would be perfect and countable, which is impossible.
This set surprised me: I didn't know we can have a no-where dense countable set such that every member of the set is a l.p. of the set. But there you go.
I don't know how you prove that any attempted construction of $X$ like this must have uncountably many limit points not in $X$ without alluding to the fact that perfect sets are uncountable, but maybe that is a task for another day.
